Sometimes, you are asked to calculate the combination or permutation modulo a number, for example $$$^nC_k \mod p$$$. Here I want to write about a complete method to solve such problems with a good time complexity because it took me a lot of googling and asking to finally have the complete approach. I hope this blog can help other users and save their time when they solve combinatorics problem in Codeforces.

#### Example Problem

Find the value of $$$^nC_k$$$, ($$$1 \leq n,k \leq 10^6$$$). As this number can be rather large, print the answer modulo $$$p$$$. ($$$p = 1000000007 = 10^9 + 7$$$)

#### Combination (binomial coefficients)

$$$^nC_k$$$ means how many ways you can choose $$$k$$$ items from an array of $$$n$$$ items, also denoted as $$$\binom{n}{k}$$$. This is also known as binomial coefficients. The formula for combination is

Sometimes, the denominator $$$k!(n-k)!$$$ is very large, but we can't modulo it since modulo operations can't be done independently on the denominator. $$$\frac{n!}{k!(n-k)!} \mod p \neq \frac{n! \mod p}{k!(n-k)! \mod p}$$$. Now I will introduce the modular multiplicative inverse to solve this problem.

#### Modular multiplicative inverse

The modular multiplicative inverse $$$x$$$ of $$$a$$$ modulo $$$p$$$ is defined as

Here, I will replace $$$x$$$ with $$$\text{inv}(a)$$$, so we have

Getting back to the formula for combination, we can rearrange so that

Here, we can use $$$\text{inv}(a)$$$ as follows

Now we can distribute the modulo to each of the terms by the distributive properties of modulo

Now I will discuss on how to calculate $$$\text{inv}(a)$$$

#### Fermat's Little Theorem

You can easily remember this theorem. Let $$$a$$$ be an integer and $$$p$$$ be a prime number,

It is helpful to know that the $$$p$$$ in the problem ($$$10^9 + 7$$$) is indeed a prime number! We can rearrange the equation to get

Looking back at our equation for $$$\text{inv}(a)$$$, both equations equate to 1, so we can equate them as

We can rearrange the equation to get

We now have a direct formula for $$$\text{inv}(a)$$$. However, we cannot use the `pow()`

function to calculate $$$a^{p-2}$$$ because $$$a$$$ and $$$p$$$ is a large number (Remember $$$1 \leq n,k \leq 10^6$$$) ($$$p = 10^9 + 7$$$). Fortunately, we can solve this using modular exponentiation.

#### Modular Exponentiation

To prevent integer overflow, we can carry out modulo operations **during** the evaluation of our new power function. But instead of using a while loop to calculate $$$a^{p-2}$$$ in $$$O(p)$$$, we can use a special trick called **exponentiation by squaring**. Note that if $$$b$$$ is an even number

Every time we calculate $$$a^2$$$, we reduce the exponent by a factor of 2. We can do this repeatedly until the exponent becomes zero where we stop the loop. This will give us a time complexity of $$$O(\log p)$$$ to calculate $$$a^{p-2}$$$ because we halve the exponent in each step. For the case when $$$b$$$ is odd, we can use the property

We then store the trailing $$$a$$$ into a variable. Then $$$b-1$$$ is even and we can proceed as previously stated. We can repeatedly apply these two equations to calculate $$$a^{p-2}$$$. Here I will show you the implementation of this modified `powmod()`

function to include modulo operations. `ll`

is defined as `long long`

```
ll powmod(ll a, ll b, ll p){
a %= p;
if (a == 0) return 0;
ll product = 1;
while(b > 0){
if (b&1){ // you can also use b % 2 == 1
product *= a;
product %= p;
--b;
}
a *= a;
a %= p;
b /= 2; // you can also use b >> 1
}
return product;
}
```

Then we can finally implement the $$$\text{inv}(a)$$$ function simply as

```
ll inv(ll a, ll p){
return powmod(a, p-2, p);
}
```

Then, finally, we can implement $$$^nC_k$$$ as

```
ll nCk(ll n, ll k, ll p){
return ((fact[n] * inv(fact[k], p) % p) * inv(fact[n-k], p)) % p;
}
```

We used the dp-approach for factorial where the factorial from 1 to n is pre-computed and stored in an array `fact[]`

.

#### Time complexity

- Pre-computation of factorial: $$$O(n)$$$
- Calculation of $$$^nC_k$$$, which is dominated by modular exponentiation
`powmod`

: $$$O(\log p)$$$ - Total: $$$O(n + \log p)$$$

#### Reference

- Fermat's Little Theorem — wiki
- Modular Multiplicative Inverse — wiki
- Modular Multiplicative Inverse — cp-algorithm

#### Problems for you

- 300C - Beautiful Numbers (Example solution: 83862274)
- 717A - Festival Organization

Please comment below if you know similar problems.

I hope this blog will help you in your competitive programming journey.

Stay safe and thank you for reading.

Auto comment: topic has been updated by jeqcho (previous revision, new revision, compare).Good tutorial! It's worth mentioning that you can find modular inverse using Extended Euclidean Algorithm in $$$O(\log{p})$$$, too.

Also, if $$$n$$$ and $$$k$$$ are small, you can calculate binomial coefficients with DP in $$$O(nk)$$$ without modular inverse.

And here are some problems:

`// you can also use (b >>= 1)`

.for correction.

Thank you. This tutorial was very helpful and easy to understand.

Thank you for tutorial!